Question

rewrite sin^2(x)cos^4(x)=(1/8)[1+cos(2x)-cos^2(2x)-cos^3(2x)]

Answer 1

\(\large \begin{align} \color{black}{sin^2xcos^4x\\~\\
=\dfrac{(2sinxcosx)^2\times cos^2x}{4}\\~\\
=\dfrac{(sin(2x))^2\times 2cos^2x}{8}\\~\\
=\dfrac{(sin(2x))^2\times (cos^2x+cos^2x)}{8}\\~\\
=\dfrac{(sin(2x))^2\times (1-sin^2x+cos^2x)}{8}\\~\\
=\dfrac{(sin(2x))^2\times (1+cos(2x))}{8}\\~\\
=\dfrac{sin^2(2x)+sin^2(2x)cos(2x)}{8}\\~\\
=\dfrac{sin^2(2x)+(1-cos^2(2x))cos(2x)}{8}\\~\\
=\dfrac{sin^2(2x)+cos(2x)-cos^3(2x)}{8}\\~\\
=\dfrac{1-cos^2x+cos(2x)-cos^3(2x)}{8}\\~\\
=\dfrac{1+cos(2x)-cos^2x-cos^3(2x)}{8}}\end{align}\)